Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin x - \cos x = \sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right)\\
{\sin ^2}x + {\cos ^2}x = 1\\
1,\\
\sin 2x - 4\left( {\cos x - \sin x} \right) = 4\\
\Leftrightarrow 2\sin x.\cos x - 4\left( {\cos x - \sin x} \right) - 4 = 0\\
\Leftrightarrow \left( { - 1 + 2\sin x.\cos x} \right) - 4.\left( {\cos x - \sin x} \right) - 3 = 0\\
\Leftrightarrow - \left( {1 - 2\sin x.\cos x} \right) - 4.\left( {\cos x - \sin x} \right) - 3 = 0\\
\Leftrightarrow - \left( {{{\sin }^2}x + {{\cos }^2}x - 2\sin x.\cos x} \right) - 4\left( {\cos x - \sin x} \right) - 3 = 0\\
\Leftrightarrow - {\left( {\sin x - \cos x} \right)^2} + 4.\left( {\sin x - \cos x} \right) - 3 = 0\\
\Leftrightarrow {\left( {\sin x - \cos x} \right)^2} - 4\left( {\sin x - \cos x} \right) + 3 = 0\\
\Leftrightarrow \left[ {\left( {\sin x - \cos x} \right) - 1} \right].\left[ {\left( {\sin x - \cos x} \right) - 3} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {\sin x - \cos x} \right) - 1 = 0\\
\left( {\sin x - \cos x} \right) - 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x - \cos x = 1\\
\sin x - \cos x = 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = 1\\
\sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\\
\sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{3}{{\sqrt 2 }}
\end{array} \right.\\
- 1 \le \sin \left( {x - \dfrac{\pi }{4}} \right) \le 1 \Rightarrow \sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
x - \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
3,\\
\left( {1 - \sqrt 2 } \right)\left( {1 + \sin x - \cos x} \right) = \sin 2x\\
\Leftrightarrow \left( {1 - \sqrt 2 } \right) + \left( {1 - \sqrt 2 } \right).\left( {\sin x - \cos x} \right) - \sin 2x = 0\\
\Leftrightarrow \left( {{{\sin }^2}x + {{\cos }^2}x} \right) - \sqrt 2 + \left( {1 - \sqrt 2 } \right).\left( {\sin x - \cos x} \right) - 2\sin x.\cos x = 0\\
\Leftrightarrow \left( {{{\sin }^2}x + {{\cos }^2}x - 2\sin x.\cos x} \right) + \left( {1 - \sqrt 2 } \right).\left( {\sin x - \cos x} \right) - \sqrt 2 = 0\\
\Leftrightarrow {\left( {\sin x - \cos x} \right)^2} + \left( {1 - \sqrt 2 } \right).\left( {\sin x - \cos x} \right) - \sqrt 2 = 0\\
\Leftrightarrow {\left( {\sin x - \cos x} \right)^2} + \left( {\sin x - \cos x} \right) - \sqrt 2 .\left( {\sin x - \cos x} \right) - \sqrt 2 = 0\\
\Leftrightarrow \left( {\sin x - \cos x} \right).\left[ {\left( {\sin x - \cos x} \right) + 1} \right] - \sqrt 2 .\left[ {\left( {\sin x - \cos x} \right) + 1} \right] = 0\\
\Leftrightarrow \left( {\sin x - \cos x + 1} \right).\left( {\sin x - \cos x - \sqrt 2 } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x - \cos x = - 1\\
\sin x - \cos x = \sqrt 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = - 1\\
\sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = \sqrt 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x - \dfrac{\pi }{4}} \right) = - \dfrac{1}{{\sqrt 2 }}\\
\sin \left( {x - \dfrac{\pi }{4}} \right) = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{4} = - \dfrac{\pi }{4} + k2\pi \\
x - \dfrac{\pi }{4} = \dfrac{{5\pi }}{4} + k2\pi \\
x - \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{{3\pi }}{2} + k2\pi \\
x = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
5,\\
\sin 2x + \sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = 1\\
\Leftrightarrow 1 - \sin 2x - \sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = 0\\
\Leftrightarrow \left( {{{\sin }^2}x + {{\cos }^2}x} \right) - 2.\sin x.\cos x - \sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = 0\\
\Leftrightarrow \left( {{{\sin }^2}x - 2\sin x.\cos x + {{\cos }^2}x} \right) - \sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = 0\\
\Leftrightarrow {\left( {\sin x - \cos x} \right)^2} - \sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = 0\\
\Leftrightarrow {\left[ {\sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right)} \right]^2} - \sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = 0\\
\Leftrightarrow 2.{\sin ^2}\left( {x - \dfrac{\pi }{4}} \right) - \sqrt 2 .\sin \left( {x - \dfrac{\pi }{4}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x - \dfrac{\pi }{4}} \right) = 0\\
\sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{4} = k\pi \\
x - \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
x - \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{2} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
